# Interesting problem on “neighbor fractions”

This is from I. M. Gelfand's Algebra book.

Fractions $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$ are called neighbor fractions if their difference $\displaystyle\frac{ad - bc}{bd}$ has numerator of $\pm 1$, that is, $ad - bc = \pm 1$. Prove that

(a) in this case neither fraction can be simplified (that is, neither has any common factors in numerator and denominator);

(b) if $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$ are neighbor fractions, then $\displaystyle\frac{a+c}{b+d}$ is between them and is a neighbor fraction for both $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$; moreover,

(c) no fraction $\displaystyle\frac{e}{f}$ with positive integer $e$ and $f$ such that $f < b + d$ is between $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$.

Parts (a) and (b) weren't too difficult, but I'm stuck on part (c). I've included (a) and (b) in case they're related to the solution to (c).

• Look up "Farey sequence". – Yuval Filmus Jun 5 '11 at 19:05
• Okay, I'll do that. Thanks! – Matt Gregory Jun 5 '11 at 19:09

## 1 Answer

Assume $\frac{e}{f}$ is (strictly) between $\frac{a}{b}$ and $\frac{c}{d}$. Then $\left|\frac{a}{b}-\frac{e}{f}\right| + \left|\frac{e}{f}-\frac{c}{d}\right| = \left|\frac{a}{b}-\frac{c}{d}\right| = \frac{1}{bd}$

But $\left|\frac{a}{b}-\frac{e}{f}\right| \geq \frac{1}{bf}$ and $\left|\frac{e}{f}-\frac{c}{d}\right|\geq \frac{1}{df}$. So $\frac{1}{bf} + \frac{1}{df} \leq \frac{1}{bd}$. Multiply both sides by $bdf$ and we get that $b+d\leq f$.

• There is a geometric meaning to this, which is related to lattice points on the plane. If $(a,b)$ and $(c,d)$ are points on the plane, then $|ad-bc|$ is the area of the quadrilateral with points $(0,0)$, $(a,b)$, $(c,d)$ and $(a+c,b+d)$. If $a,b,c,d$ are integers, and $|ad-bc| = 1$, this theorem essentially says that there is no lattice point in the inside of the quadrilateral. – Thomas Andrews Jun 5 '11 at 20:26
• To be clear, what does "between" mean in this context? Can $\frac{e}{f}$ be equal to one of the other fractions? – MathematicsStudent1122 Jul 8 '15 at 7:49
• In this case, I wrote "strictly' precisely because I wanted to be clear that I meant $\frac{a}{b}<\frac{e}{f}<\frac{c}{d}$ or $\frac{c}{d}<\frac{e}{f}<\frac{a}{b}$. @ArshJhaj The theorem is not true if you allow $e=c,f=d$. The first part of my line is true for $\frac{e}{f}=\frac{a}{b}$, but the next step: $$\left|\frac{a}{b}-\frac{e}{f}\right|\geq \frac{1}{bf}$$ is obviously not – Thomas Andrews Jul 8 '15 at 10:59